Question

Let $c$ be boundary of $[0, 1] \times [0, 1]$ oriented counter clockwise then $\int_c (y^4 + x^3) dx + 2 x^6 dy$ is

• A. 1
• B. 0
• C. 12
• D. 2

Hint:

Green's Theorem transforms tedious 4-segment line integrals along the boundary of a rectangle into a straightforward 2D double integral!

Answer 1

The Correct Option is A

Step 1: Concept
Green's Theorem in the Plane: Let $C$ be a positively oriented (counterclockwise), piecewise-smooth, simple closed curve in $\mathbb{R}^2$, and let $D$ be the region enclosed by $C$. If $P(x, y)$ and $Q(x, y)$ have continuous partial derivatives on $D$: \[ \oint_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy \]

Step 2: Key Formulas and Approach
Here $P(x, y) = y^4 + x^3$ and $Q(x, y) = 2 x^6$. The region $D$ is the unit square $[0, 1] \times [0, 1]$ in the $xy$-plane. Compute partial derivatives $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$, then evaluate the double integral over $D$.

Step 3: Step-by-step Explanation

• Compute $\frac{\partial Q}{\partial x}$: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2 x^6) = 12 x^5 \]
• Compute $\frac{\partial P}{\partial y}$: \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y^4 + x^3) = 4 y^3 \]
• Set up the double integral over $D = [0, 1] \times [0, 1]$: \[ I = \int_{y=0}^1 \int_{x=0}^1 (12 x^5 - 4 y^3) \, dx \, dy \]
• Integrate with respect to $x$: \[ \int_0^1 (12 x^5 - 4 y^3) \, dx = \left[ 2 x^6 - 4 x y^3 \right]_{x=0}^{x=1} = (2 - 4 y^3) - 0 = 2 - 4 y^3 \]
• Integrate with respect to $y$: \[ I = \int_0^1 (2 - 4 y^3) \, dy = \left[ 2 y - y^4 \right]_0^1 = (2(1) - 1^4) - 0 = 2 - 1 = 1 \]

Step 4: Final Answer
The line integral evaluates to 1. Thus, Option (A) is correct.